Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $r = \dfrac{-4t^2 - 60t - 216}{3t^2 + 30t + 27} \div \dfrac{-4t^2 - 24t}{-4t^2 - 16t} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{-4t^2 - 60t - 216}{3t^2 + 30t + 27} \times \dfrac{-4t^2 - 16t}{-4t^2 - 24t} $ First factor out any common factors. $r = \dfrac{-4(t^2 + 15t + 54)}{3(t^2 + 10t + 9)} \times \dfrac{-4t(t + 4)}{-4t(t + 6)} $ Then factor the quadratic expressions. $r = \dfrac {-4(t + 9)(t + 6)} {3(t + 9)(t + 1)} \times \dfrac {-4t(t + 4)} {-4t(t + 6)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac { -4(t + 9)(t + 6) \times -4t(t + 4)} { 3(t + 9)(t + 1) \times -4t(t + 6)} $ $r = \dfrac {16t(t + 9)(t + 6)(t + 4)} {-12t(t + 9)(t + 1)(t + 6)} $ Notice that $(t + 9)$ and $(t + 6)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {16t\cancel{(t + 9)}(t + 6)(t + 4)} {-12t\cancel{(t + 9)}(t + 1)(t + 6)} $ We are dividing by $t + 9$ , so $t + 9 \neq 0$ Therefore, $t \neq -9$ $r = \dfrac {16t\cancel{(t + 9)}\cancel{(t + 6)}(t + 4)} {-12t\cancel{(t + 9)}(t + 1)\cancel{(t + 6)}} $ We are dividing by $t + 6$ , so $t + 6 \neq 0$ Therefore, $t \neq -6$ $r = \dfrac {16t(t + 4)} {-12t(t + 1)} $ $ r = \dfrac{-4(t + 4)}{3(t + 1)}; t \neq -9; t \neq -6 $